If x is a positive integer, what is the units digit of 24^(5 + 2x)*36^

This question can be solved quickly by finding the cycles of each of the numbers. I will leave 24 for last.

36 - the units digit for 36^n will always end in 6. Why? Because 6 * 6 = 36. The cycle continues.

17 - we are only worried about the units digit so 7x7 = 49. That takes care of 17^2
for 17^3, we want 9*7 = 63
We see that 17^3 ends in 3


Now looking at 24.

cycle 1: 4*4 = 16 ends in 6
cycle 2: 6*4 = 24
cycle 3: 4*4 = 16 so know we see that we cycle between 6 and 4 depending on whether the power is even or odd

The power for 24 is 5+2x, which will always be odd (O+E=O). Using the cycle above odd's end in 4.


Now that we have our 3 endings, we multiple them together. 4*6*3 = 24*3= 72


24^{5 + 2x}*36^6*17^3 units digit will be 2. Option A

Convert it into the power of primes
2^27 * 2^6x * 3^17 * 17^3
using theconcept of cyclicity
units digit would be
8*4*7*3= 2(units digit of the given expression)

MANHATTAN GMAT OFFICIAL SOLUTION:

a. Unknowns: Units digits of 24^(5 + 2x), 36^6, and 17^3.
Given: We only need the units digit of the product, not the value. x is a positive integer.
Constraints: If x is a positive integer, 2x is even, and 5 + 2x must be odd. Units digit of a product depends only on the units digit of multiplied numbers.
Question: What is the units digit of the product 24^(5 + 2x)*36^6*17^3?

b. Find/Recall the pattern for units digits → use the Unit Digit Shortcut

c. Units digit of 24^(5 + 2x) = units digit of 4^odd. The pattern for the units digit of 4^(integer) = [4, 6]. Thus, the units digit is 4.
Units digit of 36^6 must be 6, as every power of 6 ends in 6.
Units digit of 1^3 = units digit of 7^3. The pattern for the units digit of 7 integer = [7, 9, 3, 1]. Thus, the units digit is 3.
The product of the units digits is (4)(6)(3) = 72, which has a units digit of 2. The answer is A.

d. Patterns were very important on this one! If we had forgotten any of the patterns, we could just list at least the first four powers of 4, 6, and 7 to recreate them.

Hi Farina

From the solution provided above by Bunuel:

- The pattern for the units digit of 4^(integer) = [4, 6] -- means any power of 4, for e.g. 4^1 = 4, 4^2 = 16(unit digit 6), 4^3 = 64(unit digit 4)....so on.
- If the power of 4 is odd(4^odd) number: for e.g. 4^1 = 4, 4^3 = 64(unit digit 4), 4^5 = 1024(unit digit 4)...so on.

- Notice that unit digit(4) is same for all 4^odd.

Hope it is clear.

Cyclicity for 24
24^1=24
24^2=576 and so on
So, 5+2x will be either 24^7 or 24^11 and so on
Therefore last digit of 24*5+2x=4

And, last digit of 36^6=6
Last digit of 17^3=3
Therefore, last digit of the question=4*6*3=2
Answer A

This is how I tried...

Cyclicity of 6 is 1 and we get 6 as unit digit for every power.
cyclicity of 7 is 4 and we get 3 as unit digit for power 3.

Now cyclicity of 4 is 2 and we get 4 or 6 depending upon the power , if power is 1 we get 4 and if power is 2 we get 6 ...similarly...for remaining

Let's try with 5+2x = 1 => 2x => -1 = > x =-1/2...this is not an integer.
if 5+2x = 2 => 2x = -3 => x = -3/2...
...
5+2x = 5 => x = 0...we get first integer ( but 0 is neither positive or negative)

5+2x = 7 => x = 1

then 4^5+2(1) => 4^7 = we get 4 as the unit's digit.

The product is 4*6*3 = 24*3 = 2 is unit's digit.

I would like to see different solutions...

Hi Bunuel,

Can you please explain this line:

The pattern for the units digit of 4^(integer) = [4, 6]. Thus, the units digit is 4.

We need to find the units digit of \((24)^{5+2x}\) * \((36)^6\) * \((17)^3\)

Now, Units digit of \((24)^{5+2x}\) will be same as units digit of \((4)^{5+2x}\) (As units digit of 24 is same as 4 which is 4)
Units digit of \((36)^6\) will be same as units digit of \((6)^6\) (As units digit of 36 is same as 6 which is 6)
Units digit of \((17)^3\) will be same as units digit of \((7)^3\) (As units digit of 17 is same as 7 which is 7)

Let's find the units digit of \((4)^{5+2x}\)

To find the units digit of power of 4 we need to check the cyclicity in the units digit of powers of 4

\(4^1\) units’ digit is 4 [ 4 ]
\(4^2\) units’ digit is 6 [ 16 ]
\(4^3\) units’ digit is 4 [ 64 ]
\(4^4\) units’ digit is 6 [ 256 ]

=> Units digit of any positive odd integer power of 4 is 4
=> Units digit of any positive even integer power of 4 is 6
=> Now 2x + 5 will be Evan + Odd = Odd (As 2x is Even and 5 is odd)
=> Units digit of \((4)^{5+2x}\) = 4

Let's find the units digit of \((6)^6\)

To find the units digit of power of 6 we need to check the cyclicity in the units digit of powers of 6

\(6^1\) units’ digit is 6 [ 6 ]
\(6^2\) units’ digit is 3 [ 36 ]
\(6^3\) units’ digit is 3 [ 216 ]

=> Units digit of any positive integer power of 6 is 6
=> The units digit of \(6^6\) = 6

Let's find the units digit of \((7)^3\)

\(7^1\) units’ digit is 7 [ 7 ]
\(7^2\) units’ digit is 9 [ 7*7 = 49 ]
\(7^3\) units’ digit is 3 [ 9*7 = 63 ] (P.S>: we don't need to proceed further)
\(7^4\) units’ digit is 1 [ 3*7 = 21 ]
\(7^5\) units’ digit is 3 [ 1*7 = 7 ]

=> Units digit of power of 7 repeats after every \(4^th\) number

=> Units digit of \((24)^{5+2x}\) * \((36)^6\) * \((17)^3\) = Units digit of 4 * 6 * 3 = Units digit of 24*3 = 2

So, Answer will be A.
Hope it helps!

Watch the following video (from 2:48 mins) to learn how to find cyclicity of 3 and other numbers

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